14 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Think about it this way:
What is the last element of the second row? (1+2)
What about the third row? (1+2+3) as 1 element in the first row, 2 in the second and 3 in the third
So, build upon this and
1. The last element is n(n+1)/2, for some n belonging to N.
2. If 5310 belongs in range [n(n-1)/2+1, n(n+1)/2], what should be the value of n?
3. This value of n gives us the row this is in.
Is the answer 103?
2 not understood what
yus
We've established that the last element of nth row is n(n+1)/2, so the last element of the previous row is n(n-1)/2 so the first element of the nth row is n(n-1)/2+1 right?
ohhhhhh okkk
i see i see
damn how did u even think of that
SirLance 🙏🙏🙏🙏🙏
This was basically the buildup to the line of thought of "can I set some upper and lower bound to the elements of a row?" 'cuz if you have the bounds, you can get the row you need. Now for the last elements, the number of terms before it is 1+2+...(n-1) from previous rows and n-1 from current row. So summing these and adding one you get the last term. (n(n+1)/2)
this is the final inequality right ?
yeah ig its called method of difference
Precisely.
Not quite the same, considering what's happening here.
eh yeah but you can easily recognise the pattern 1,3,6,10
can't you use tn-tn-1?
each term is the last element of each row
t1=1 t2=3 t3=6 t4=10
the difference has a difference of 1
I mean you'll get the same result anyway but this is just another form of understanding how SirLancelot thought of the idea